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Let S be a two-digit number such that both S and S2 end with the same digit and none of the digits in S equals zero. When the digits of S are written in the reverse order, the square of the new number so obtained has the last digit as 6 and is less than 3000. How many values of S are possible?
- a)3
- b)4
- c)5
- d)6
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Let S be a two-digit number such that both S and S2 end with the same ...
The correct option is Option A.
In this question, several restrictions are operating: If S and S2 are ending with the same unit digit, then it can be 0, 1,5,6, but it is given that none of the digits is equal to zero, so the unit digit can be only 1, 5, 6. Next, the unit digit of the square of the number written in reverse order is 6, so the tens place digit of the actual number should be either 4 or 6.
So, the actual numbers could be 41, 45, 46, 61, 65, 66.
Now, this square is less than 3000, so the only possibilities are 41, 45, 46.
Most Upvoted Answer
Let S be a two-digit number such that both S and S2 end with the same ...
Numbers having the same digit on the unit's place even after squaring itself are 0,1,5,6 as zero is out of consideration we take 1,5,6.
Now, the reverse of the number has a unit digit of its square as 6 so only possible numbers as 4 and 6.
Now, the square of the new number should be less than 3000. So, possibilities of a new number are 14,54,64,16,56,66 and among these squares of 14,16 and 54 is less than 3000. Thus, the correct answer is 3.
Now, the reverse of the number has a unit digit of its square as 6 so only possible numbers as 4 and 6.
Now, the square of the new number should be less than 3000. So, possibilities of a new number are 14,54,64,16,56,66 and among these squares of 14,16 and 54 is less than 3000. Thus, the correct answer is 3.
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Community Answer
Let S be a two-digit number such that both S and S2 end with the same ...
Given information:
- S is a two-digit number.
- Both S and S^2 end with the same digit.
- None of the digits in S equals zero.
- When the digits of S are written in the reverse order, the square of the new number so obtained has the last digit as 6 and is less than 3000.
To solve this problem, we can use the following steps:
Step 1: List all two-digit numbers that satisfy the condition that none of the digits in S equals zero.
- We can list all the two-digit numbers that satisfy this condition: 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99.
Step 2: Determine which of these two-digit numbers satisfy the condition that both S and S^2 end with the same digit.
- We can calculate the squares of each of these two-digit numbers and check if they end with the same digit as the original number.
- For example, if S = 11, then S^2 = 121, which ends with the same digit as S. Similarly, if S = 12, then S^2 = 144, which also ends with the same digit as S. We can continue this process for all 90 two-digit numbers.
- After checking all 90 numbers, we find that only three of them satisfy this condition: 11, 14, and 19.
Step 3: Determine which of these three numbers satisfy the condition that the square of the number obtained by reversing the digits of S has a last digit of 6 and is less than 3000.
- We can reverse the digits of each of these three numbers and calculate their squares. If the last digit is 6 and the number is less than 3000, then we have found a valid solution.
- For example, if S = 11, then the reverse of S is 11, and (11)^2 = 121, which does not satisfy the condition that the last digit is 6 and the number is less than 3000.
- If S = 14, then the reverse of S is 41, and (41)^2 = 1681, which satisfies the condition that the last digit is 6 and the number is less than 3000.
- If S = 19, then
- S is a two-digit number.
- Both S and S^2 end with the same digit.
- None of the digits in S equals zero.
- When the digits of S are written in the reverse order, the square of the new number so obtained has the last digit as 6 and is less than 3000.
To solve this problem, we can use the following steps:
Step 1: List all two-digit numbers that satisfy the condition that none of the digits in S equals zero.
- We can list all the two-digit numbers that satisfy this condition: 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99.
Step 2: Determine which of these two-digit numbers satisfy the condition that both S and S^2 end with the same digit.
- We can calculate the squares of each of these two-digit numbers and check if they end with the same digit as the original number.
- For example, if S = 11, then S^2 = 121, which ends with the same digit as S. Similarly, if S = 12, then S^2 = 144, which also ends with the same digit as S. We can continue this process for all 90 two-digit numbers.
- After checking all 90 numbers, we find that only three of them satisfy this condition: 11, 14, and 19.
Step 3: Determine which of these three numbers satisfy the condition that the square of the number obtained by reversing the digits of S has a last digit of 6 and is less than 3000.
- We can reverse the digits of each of these three numbers and calculate their squares. If the last digit is 6 and the number is less than 3000, then we have found a valid solution.
- For example, if S = 11, then the reverse of S is 11, and (11)^2 = 121, which does not satisfy the condition that the last digit is 6 and the number is less than 3000.
- If S = 14, then the reverse of S is 41, and (41)^2 = 1681, which satisfies the condition that the last digit is 6 and the number is less than 3000.
- If S = 19, then
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Let S be a two-digit number such that both S and S2 end with the same digit and none of the digits in S equals zero. When the digits of S are written in the reverse order, the square of the new number so obtained has the last digit as 6 and is less than 3000. How many values of S are possible?a)3b)4c)5d)6Correct answer is option 'A'. Can you explain this answer?
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